Question: Let $a$ and $b$ be real numbers so that the roots of
\[z^2 + (10 + ai) z + (27 + bi) = 0\]are complex conjugates.  Enter the ordered pair $(a,b).$
Solution: Let $z = x + yi,$ where $x$ and $y$ are real numbers.  Then both
\[z + \overline{z} = (x + yi) + (x - yi) = 2x,\]and
\[z \overline{z} = (x + yi)(x - yi) = x^2 + y^2\]are real numbers.  Therefore, by Vieta's formulas, all the coefficients must be real numbers.  Then $(a,b) = \boxed{(0,0)}.$